A box can be formed by cutting a square out of each corner of a piece of cardboard and folding the sides up. If the piece of cardboard is 72 cm by 72 cm and each side of the square that is cut out has length x​ cm, the function that gives the volume of the box is Vequals5184xminus288x squaredplus4 x cubed. Complete parts​ (a) and​ (b) below.

Question

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Part 1
x
x
72
72
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Part 1
a. Find the values of x that make Vgreater than0. Select the correct choice below and fill in any answer boxes to complete your choice.
A.
The values of x that make Vgreater than0 are the interval
  
enter your response here.
​(Type your answer in interval​ notation.)
B.
The values of x that make Vgreater than0 are the​ point(s)
  
enter your response here.
​(Use a comma to separate answers as​ needed.)
C.
There are no values of x that make Vgreater than0.

Answer 1

To find the values of \( x \) that make \( V > 0 \) for the volume function

\[ V = 5184x - 288x^2 + 4x^3, \]

we need to analyze when this cubic function is greater than zero.

Step 1: Find the roots of the function.

To determine where the volume is positive, we first need to find the roots of the function. This involves setting the function equal to zero:

\[ 5184x - 288x^2 + 4x^3 = 0. \]

This can be factored out by isolating \( x \):

\[ 4x(1296 - 72x + x^2) = 0. \]

So, one root is \( x = 0 \). Now, we need to solve the quadratic \( 1296 - 72x + x^2 = 0 \).

Using the quadratic formula, where \( a = 1, b = -72, c = 1296 \):

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{72 \pm \sqrt{(-72)^2 - 4 \cdot 1 \cdot 1296}}{2 \cdot 1} = \frac{72 \pm \sqrt{5184 - 5184}}{2} = \frac{72 \pm 0}{2} = 36. \]

So the roots of the function are \( x = 0 \) and \( x = 36 \).

Step 2: Determine the intervals to test.

The cubic function \( V(x) \) is continuous, and we can analyze its behavior over the intervals determined by its roots: \( (-\infty, 0) \), \( (0, 36) \), and \( (36, \infty) \).

Step 3: Test the intervals.

• For \( x < 0 \): Choose \( x = -1 \): \[ V(-1) = 5184(-1) - 288(-1)^2 + 4(-1)^3 = -5184 - 288 - 4 < 0. \]

• For \( 0 < x < 36 \): Choose \( x = 1 \): \[ V(1) = 5184(1) - 288(1)^2 + 4(1)^3 = 5184 - 288 + 4 = 4900 > 0. \]

• For \( x > 36 \): Choose \( x = 37 \): \[ V(37) = 5184(37) - 288(37)^2 + 4(37)^3 = 191808 - 288(1369) + 4(50653) = 191808 - 394752 + 202612 < 0. \]

Step 4: Conclusion.

Thus, the volume function \( V(x) > 0 \) for the interval \( (0, 36) \).

Answer:

The correct answer is:

A. The values of \( x \) that make \( V > 0 \) are the interval \[ \boxed{(0, 36)}. \]

B. The values of \( x \) that make \( V > 0 \) are the points \[ \text{None, as it is an interval}. \]

C. There are no values of \( x \) that make \( V > 0 \) is incorrect.

So you can submit \( A \) as your final answer.